The flux of the given function f = x³i + y³j + z³k is given by f',
f' = 3(x²i + y²j + z²k)
Given, f = x³i + y³j + z³k
we have to find the flux of the given function f = x³i + y³j + z³k
On partial differentiating the function, we get
f' = 3x²i + 3y²j + 3z²k
f' = 3(x²i + y²j + z²k)
So, the flux of the given function f = x³i + y³j + z³k is given by f',
f' = 3(x²i + y²j + z²k)
Hence, the flux of the given function f = x³i + y³j + z³k is given by f',
f' = 3(x²i + y²j + z²k)
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Answer: 3312q^2 - 9093q + 5914
Step-by-step-explanation:
Answer:
n=150
Step-by-step explanation:
(-17+n/5)-13=0
-17= -17/1 = -17(5)/5
-17(5)+n/5 = n-85/5
(n-85)/5-13=0
13=13/1 =13(5)/5
(n-85)-(13(5))/5 =n-150/5
(n-150)/5=0
n-150=0
n=150
Answer: -105
Step-by-step explanation:
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